Question

prove that =sec4a-1/sec2a-1=tan4a*cota
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Answer 1

Answer:

This is the answer

I hope can I help you

Answer 2

Answer:

sec4A -1 / sec2A-1 

 = 1/ cos4A - 1 / 1/cos2A -1

 =  (1- cos4A ) . cos2A /  cos4A (1- cos2A )

 =  2 sin2 2A . cos2A / cos4A . 2 sin2 A

 =  2 sin2A cos2A . sin2A / cos4A . 2sin2 A

 = sin4A . 2 sinA cosA / cos4A . 2 sin2 A

 =  tan4A / tanA

=tan4A. cotA

Step-by-step explanation:

[sec(4A) - 1) /[sec(2A) - 1] = tan(4A) cotA

let's rewrite sec(4A) as 1 /cos(4A), sec(2A) as 1 /cos(2A), tan(4A) as sin(4A) /cos(4A) and cotA as cosA /sinA:

{[1 /cos(4A)] - 1} /{[1 /cos(2A)] - 1} = [sin(4A) /cos(4A)] (cosA /sinA)

{[1 - cos(4A)] /cos(4A)} /{[1 - cos(2A)] /cos(2A)} =

[sin(4A) /cos(4A)] (cosA /sinA)

{[1 - cos(4A)] /cos(4A)} {cos(2A) /[1 - cos(2A)]} =

[sin(4A) /cos(4A)] (cosA /sinA)

let's apply (to both 1 - cos(4A) and 1 - cos(2A) ) the half-angle formula sin²(θ/2) = (1 - cosθ) /2 (hence 1 - cosθ = 2sin²(θ/2) ):

{{2sin²[(4A)/2]} /cos(4A)} {cos(2A) /{2sin²[(2A)/2]}} =

[sin(4A) /cos(4A)] (cosA /sinA)

{[2sin²(2A)] /cos(4A)} [cos(2A) /(2sin²A)] =

[sin(4A) /cos(4A)] (cosA /sinA)

{[2sin²(2A) cos(2A)] /cos(4A)} [1 /(2sin²A)] =

[sin(4A) /cos(4A)] (cosA /sinA)

{{[2sin(2A) cos(2A)] sin(2A)} /cos(4A)} [1 / (2sin²A)] =

[sin(4A) /cos(4A)] (cosA /sinA)

let's apply (to 2sin(2A) cos(2A) ) the double-angle formula 2sinθ cosθ = sin(2θ):

{{sin[2(2A)] sin(2A)} /cos(4A)} [1 /(2sin²A)] =

[sin(4A) /cos(4A)] (cosA /sinA)

[sin(4A) /cos(4A)] [sin(2A) /(2sin²A)] =

[sin(4A) /cos(4A)] (cosA /sinA)

let's apply again (to sin(2A) ) the double-angle forrmula sin(2θ) =

2sinθ cosθ:

[sin(4A) /cos(4A)] [(2sinA cosA) /(2sin²A)] =

[sin(4A) /cos(4A)] (cosA /sinA)

simplifying to:

[sin(4A) /cos(4A)] (cosA /sinA) = [sin(4A) /cos(4A)] (cosA /sinA) (proved)

I hope it's helpful